Why is information about the electron described with information of a photon [ ]?

1. The problem statement, all variables and given/known data
The problem and its solution are attached as TheProblemAndSolution.jpg.

2. Relevant equations
Δx = λ/sinθ
Δp_x = 2h/λ sinθ

3. The attempt at a solution
In the equation given in the problem (Δx = λ/sinθ), why is the uncertainty of the ELECTRON written as a function of the PHOTON's wavelength (and the microscope's angle)? Similarly, in the equation given in the solution (Δp_x = 2h/λ sinθ), why is the uncertainty in the x component of the momentum of the ELECTRON written, again, as a function of the PHOTON's wavelength (and, again, the microscope's angle)? If it wasn't clear from the above, I am placing emphasis on the fact that Heisenberg's Uncertainty Principle is all about being able to measure, with great certainty, either the position or momentum of the ELECTRON while the other factor remains highly uncertain BY USING CHARACTERISTIC VALUES OF A PHOTON (RATHER THAN AN ELECTRON). I would really appreciate it if someone could tell me why this is the case.

P.S.
The involvement of the sine of the angle of the microscope is not baffling or confusing to me, it's that I don't understand why information about the photon is used to describe uncertainty of a characteristic of an electron.

P.P.S
I am not looking for an overly complex answer (unless it is inherently so) so, please keep the answer as simple as possible.

Re: Why is information about the electron described with information of a photon [...

This is not homework. (I don't know if you noticed but, the solution is included in my main post.)

I'm just trying to understand this problem inside-out. I'd appreciate it if I can have an answer that simply addresses my concerns and nothing more (in order to not confuse me or add unnecessary difficulty).

Re: Why is information about the electron described with information of a photon [...

the size of an object which you want to take a picture of must be at least as large as the wavelength of light which you use to take the picture. This is a law of optics. (You can work it out using classical wave theory). Therefore, the uncertainty in the electron's position will scale with the wavelength of the photon.

And the uncertainty of the momentum of the electron is the same as the uncertainty in the momentum of the photon. This is because momentum is conserved in the collision, so if the photon had an uncertainty in its momentum, then after the collision, the electron will have an uncertainty in its momentum.

Re: Why is information about the electron described with information of a photon [...

Therefore, the uncertainty in the electron's position will scale with the wavelength of the photon.

It makes sense to me when you said that “the size of an object which you want to take a picture of must be at least as large as the wavelength of light which you use to take the picture.” but, could you help me intuitively understand where the Δx = λ/sinθ equation comes from?

This is because momentum is conserved in the collision

Could you please also explain why the conservation of momentum in the collision implies that the uncertainty of the momentum and the photon are equivalent?

Re: Why is information about the electron described with information of a photon [...

Therefore, the uncertainty in the electron's position will scale with the wavelength of the photon.

It makes sense to me when you said that “the size of an object which you want to take a picture of must be at least as large as the wavelength of light which you use to take the picture.” but, could you please help me intuitively understand where the entire Δx = λ/sinθ equation comes from?

This is because momentum is conserved in the collision.

Could you please also explain why the conservation of momentum in the collision implies that the uncertainty of the momentum of the electron and that of the photon are equivalent?

It makes sense to me when you said that “the size of an object which you want to take a picture of must be at least as large as the wavelength of light which you use to take the picture.” but, could you help me intuitively understand where the Δx = λ/sinθ equation comes from?

It is simple singe-slit diffraction. I.e. if the light comes from a place of width d, then it will end up moving in a direction where theta is less than the theta given by the formula (most of the time). As I said, this is classical physics. So the question is kind of using a mix of classical and quantum reasoning. This question is not doing the calculations very precisely, the point of the question is to give you a feeling of the uncertainty principle, not working knowledge of it.

s3a said:

Could you please also explain why the conservation of momentum in the collision implies that the uncertainty of the momentum and the photon are equivalent?

Well, I am assuming it is the photon which initially has most of the momentum in the system, and its momentum is initially well-defined, so the total momentum of the system is well-defined. And then after collision, we know that the momentum of the photon and electron must add up to this initial value. We can repeat this experiment many times to find out how the photon's momentum after the collision varies from its average. And we know that each time we do this experiment, the momentum of the electron + that of the photon must add up to a constant value. Therefore, the way the electron's momentum varies over many experiments must be exactly the same as the way the photon's momentum varies. Therefore, the uncertainty in their momentum must be the same.

EDIT: in other words, imagine doing 10 experiments. For each experiment, you will obtain different values for the final momentum of the photon. We know momentum is conserved for each experiment, so we can write down the final momentum of the electron in each experiment. If we calculate the spread in the values of electron and photon momentum, then hey-presto, they will be the same, because in each experiment, the electron momentum is simply the same constant (initial momentum), minus the photon momentum. We know that adding the same constant to each data point doesn't change the spread of the values, therefore the spread is the same.

Re: Why is information about the electron described with information of a photon [...

I am not satisfied in “just getting” the idea that you can only know either the momentum or the position of the electron with great certainty. Not knowing everything I need to know to make intuitive sense of it ultimately confuses me.

(1) The formula for the uncertainty of the electron's position is starting to make more sense to me now. It is assuming m = 1 (first order), right?

(2) You say that the total momentum is well-defined because the initial momentum of the photon is well-defined (and I assuming you're just taking the word of the problem for the electron's momentum being well-defined). But, how, exactly, is the electron's momentum well-defined?

(3) What logic or evidence is your assumption about the photon always having a greater initial momentum than the electron based on? Does this even matter? If it is the case, that the electron's initial momentum is larger, everything you said about the uncertainties of their momenta being the same still stands, right?

(4) Despite “just knowing” that a greater uncertainty about the momentum of the electron or its position will imply a smaller uncertainty in the other factor, I don't see how the position is related to momentum in the first place. I get that momentum is the product of mass times velocity and the velocity is the rate of change of displacement which is a net change in position but, beyond that, I don't see how the two are related. What's their “initial relationship” such that Heisenberg could even think about writing down that Δx Δp_x = constant (or an inequality instead of the equality)?

Also, I now get how their uncertainty in momentum is the same (thanks to you). :)

If anything I said is unclear, just ask me to rephrase it in a better way.

I am not satisfied in “just getting” the idea that you can only know either the momentum or the position of the electron with great certainty. Not knowing everything I need to know to make intuitive sense of it ultimately confuses me.

Yep, for the same reason, this problem confused me a lot when I first saw it. It is probably best to try not to think on this problem for too long. It is best to start learning quantum mechanics. (for example, particle in a square well, particle in free space, hydrogen atom, etc. These problems are better to get an understanding of the basics of quantum mechanics).

s3a said:

(1) The formula for the uncertainty of the electron's position is starting to make more sense to me now. It is assuming m = 1 (first order), right?

I'm not sure what you mean by this.

s3a said:

(2) You say that the total momentum is well-defined because the initial momentum of the photon is well-defined (and I assuming you're just taking the word of the problem for the electron's momentum being well-defined). But, how, exactly, is the electron's momentum well-defined?

(3) What logic or evidence is your assumption about the photon always having a greater initial momentum than the electron based on? Does this even matter? If it is the case, that the electron's initial momentum is larger, everything you said about the uncertainties of their momenta being the same still stands, right?

Hmm. The reason that the question assumes the initial momentum of the electron is well defined is to make the problem as simple as possible. This can be done by making a beam of electrons which is sufficiently spread out throughout space. And no, in this problem, it doesn't matter which particle has the most momentum.

s3a said:

(4) Despite “just knowing” that a greater uncertainty about the momentum of the electron or its position will imply a smaller uncertainty in the other factor, I don't see how the position is related to momentum in the first place. I get that momentum is the product of mass times velocity and the velocity is the rate of change of displacement which is a net change in position but, beyond that, I don't see how the two are related. What's their “initial relationship” such that Heisenberg could even think about writing down that Δx Δp_x = constant (or an inequality instead of the equality)?

Excellent question. Historically, quantum mechanics came about because people started to see places where the classical theory did not hold. For example, electromagnetic radiation seemed to be explained much better if it were absorbed and emitted in discrete 'quanta'. People also related the quantum-mechanical theory to electrons, and it turned out to be a good model for them, and people started diffracting electrons, similarly to how you would diffract light. So, we have a theory which explains much of the behaviour of both electrons and photons. The uncertainty principle is really self-evident, given the postulates of quantum mechanics. In other words, since we have a theory (quantum mechanics) which describes our world very well, then for the time being we assume its postulates are not dis-proven. Therefore the uncertainty principle is not dis-proven. (You could change the phrase 'not dis-proven' with 'true', but 'truth' sounds like quite an unscientific word, right?)

EDIT: To answer your last question in another way, we know that for electromagnetic radiation, wavelength and position are intimately linked, and this wave nature (which is more obvious for electromagnetic theory) is one 'facet' of quantum theory, which is a theory underlying photons and electrons. So it is not so surprising that for electrons, wavelength and position are also intimately linked. And also, the inverse of wavelength is proportional to momentum, therefore momentum and position are linked in a special way in quantum mechanics.

I agree since we can never know for sure if a certain theory is a representation of the truth or a qualitatively good approximation of it until we find evidence disproving it.

1) How does you measure the final momentum of the photon bounced in real life? Do you analyze a single-slit diffraction image on a screen?

2) How do you measure the position of the photon? Do you at all or do you just say “I know it's somewhere in the slit” and then you attempt to make the slit as small as you can (by using light with as small a wavelength as possible) and then, since the uncertainty is small, you can say something like “screw the specifics, it's between this line here and that line there which are super close so we can just assume it's in the middle of the two lines”?

3) I get, mathematically (in the way that the solution of the problem in this thread shows), that if you decrease the wavelength of the light in an attempt to make the slit (and the electron's uncertainty in position) smaller, you will increase the uncertainty in the electron's momentum (which is what the solution of the problem in this thread is all about) however, I don't understand, scientifically, how this principle works. I know it's a fundamental limitation imposed by nature (rather than a limitation of our measurement tools). I'm looking for an answer that's on a sub-atomic particle level (that involves thinking of the collision and what happens after it).

4) In real life, do we measure the momentum of the electron that's prior to the collision by accelerating very spaced-out electrons using a potential difference and then shining light opposite to the electrons and, since light is unaffected by the potential difference due to its lack of charge, it should not introduce any new uncertainty when the collision occurs? What about the interaction of electrons between one another? That introduces new uncertainty but, is that uncertainty also part of the uncertainty principle? Do we just neglect that completely (since the interactions are minor since the electrons are spaced out heavily)?

As for the initial momentum of the light, nothing confuses me; it's simply choosing a wavelength and “shooting” it.

In short:
* I have minor doubts about how to get the initial momentum of the electron in real life that I need cleared.
* I know how to get the initial momentum of the photons in real life.
* I'm thinking that the analysis of the single-slit diffraction pattern is what allows a scientist to measure the momentum of the photon after the collision.

P.S.
Sorry if I said anything stupid or repeated myself because, I have a headache as I write this.

3) I get, mathematically (in the way that the solution of the problem in this thread shows), that if you decrease the wavelength of the light in an attempt to make the slit (and the electron's uncertainty in position) smaller, you will increase the uncertainty in the electron's momentum (which is what the solution of the problem in this thread is all about) however, I don't understand, scientifically, how this principle works. I know it's a fundamental limitation imposed by nature (rather than a limitation of our measurement tools). I'm looking for an answer that's on a sub-atomic particle level (that involves thinking of the collision and what happens after it).

This problem is fairly intuitive, if you think about the collision. (although not all problems in quantum mechanics are intuitive at all). Anyway, in this problem, If you decrease the wavelength of the light, you are increasing its frequency, and therefore increasing its momentum. When the photon collides with the electron, there is a certain probability that the electron will be 'bounced' away in a certain direction. And if the photon initially has more momentum, it will transfer more momentum to the electron.

So If we just think about what happens to the electron, after collision it gets bounced away in a random direction, and it gets bounced further if the photon had more momentum. So, after collision, the uncertainty in the momentum of the electron is greater when the photon has greater initial momentum.

1) How does you measure the final momentum of the photon bounced in real life? Do you analyze a single-slit diffraction image on a screen?

I think it depends on the type of experiment you are doing... If the light is monochrome, then you could do a simple diffraction experiment to find the wavelength (and therefore momentum). If the light is made up of lots of different wavelengths, then maybe you'd need some kind of spectrometer, for example a ccd.

2) How do you measure the position of the photon? Do you at all or do you just say “I know it's somewhere in the slit” and then you attempt to make the slit as small as you can (by using light with as small a wavelength as possible) and then, since the uncertainty is small, you can say something like “screw the specifics, it's between this line here and that line there which are super close so we can just assume it's in the middle of the two lines”?

the equation for diffraction does take into account the fact that the light could come from any point within the slit, but this is not a problem, since the slit is small compared to the distance the light travels from the slit. Does this answer your question?

4) In real life, do we measure the momentum of the electron that's prior to the collision by accelerating very spaced-out electrons using a potential difference and then shining light opposite to the electrons and, since light is unaffected by the potential difference due to its lack of charge, it should not introduce any new uncertainty when the collision occurs? What about the interaction of electrons between one another? That introduces new uncertainty but, is that uncertainty also part of the uncertainty principle? Do we just neglect that completely (since the interactions are minor since the electrons are spaced out heavily)?

We could even accelerate the electrons, let some of them come out of the 'accelerating tube' (sorry that's probably not correct terminology), then turn on the beam of photons, so that the collisions would happen without the potential difference in the background. (This would all need to be within a vacuum, of course).
And yes, we can neglect the interaction between electrons, since we can make a beam in which they are not close to one another.

Re: Why is information about the electron described with information of a photon [...

Okay, I think I'm beginning to get it more.

Right now, I need to understand two more things to my knowledge.:

1) I'm beginning to understand the Δx = λ/sinθ more. Basically, a larger wavelength takes more physical space and needs a larger slit so it can pass through. Also, sinθ will yield a value between 0 (excluded) and 1 (included) meaning that the slit will be larger or equal to the wavelength of the photon and that the photon is somewhere within the slit. In other words, it makes sense to me why Δx and λ are proportional. What doesn't make sense to me is how the sinθ part is there. I would like your help to dive into the single-slit diffraction non-formal “proof” to figure out how to get the sinθ part if you don't mind.

2) I would like to understand how the momentum uncertainty is the Δp_x = 2h/λ sinθ equation. I understand it geometrically so, what I am asking is basically when you said

We know that adding the same constant to each data point doesn't change the spread of the values, therefore the spread is the same.

, how do we know that this constant spread is (always) equal to 2h/λ sinθ? Or is that not what's going on? Assuming what I said so far is correct for #2, can you show that to me mathematically please? I believe an explanation using vector summation would be the one I am looking for.

QUOTE]p.s. sorry I've taken a while to reply, I have been under the weather.[/QUOTE]
That's fine, I'm not rusing you. In fact, I am very grateful for your answers. Also, sorry for taking long to respond myself. I've seen your post for a while now but, I was either too busy or too tired to post and it's pointless to post if I haven't thought through what I am writing (since this thread is about understanding a physical concept rather than just responding to a regular email). Technically, I am tired now too but, I'm still well-rested enough to be able to understand what's going on if I think about it slowly.